Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
Look at the other value: normal means uncompensated, abnormal in the compensating direction means partial compensation, and a normal pH with both abnormal means full compensation.
In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.
ECG rhythm strip interpretation
Paper (25 mm/s, 10 mm/mV)
1 small box = 0.04 s, 1 large box = 0.20 s, 10 mm = 1 mV
Heart rate, regular rhythm
300 ÷ large boxes = 1500 ÷ small boxes = 60 ÷ RR (s)
Heart rate, irregular rhythm
QRS complexes in 6 s × 10
Normal PR interval
0.12–0.20 s
Normal QRS duration
< 0.12 s
QTc (Bazett)
QT/√RR
QTc (Fridericia)
QT/∛RR
QTc (Framingham)
QT + 154·(1 − RR)
Sinus rates
bradycardia < 60, normal 60–100, tachycardia > 100 /min
Escape rates
junctional 40–60, ventricular 20–40 /min
Fast rhythms
SVT 150–250 /min (narrow), VT > 100 /min (wide)
Method
Rate: for a regular rhythm divide 300 by the large boxes or 1500 by the small boxes between two R waves; for an irregular rhythm count the QRS complexes in 6 seconds and multiply by 10.
Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
P waves: is there an upright P wave in lead II before every QRS, all alike? Look for fibrillatory or sawtooth waves, inverted P waves and P waves without a QRS.
PR interval: from the start of P to the start of the QRS, normally 0.12–0.20 s and constant. Watch whether it lengthens, stays fixed before a dropped beat, or varies at random.
QRS duration: below 0.12 s is narrow; 0.12 s or more is wide. Then name the rhythm, and for the QT use QTc = QT/√RR (Bazett) with times in seconds.
Evidence-based practice statistics
Sensitivity and specificity
Se = a/(a + c), Sp = d/(b + d)
Predictive values
PPV = a/(a + b), NPV = d/(c + d)
Accuracy
(a + d)/n
Likelihood ratios
LR+ = Se/(1 − Sp), LR− = (1 − Se)/Sp
PPV at prevalence p
Se·p / (Se·p + (1 − Sp)(1 − p))
Odds
odds = p/(1 − p); p = odds/(1 + odds); post-test odds = pre-test odds × LR
Risk measures
RR = EER/CER, ARR = CER − EER, RRR = ARR/CER
Number needed to treat
NNT = 1/ARR, rounded up; NNH = 1/(EER − CER)
Odds ratio
OR = ad/(bc)
Incidence rate
new cases / person-time
Method
Draw the 2 × 2 table and label a, b, c and d before you calculate.
Read sensitivity and specificity down the columns, predictive values along the rows.
For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
For a likelihood ratio, convert the probability to odds, multiply, and convert back.
For a trial, find EER and CER first; then RR = EER/CER, ARR = CER − EER and NNT = 1/ARR rounded up.
2Practice problems
1.Human anatomyMedium
Which structure produces most of the cerebrospinal fluid?
AChoroid plexus
BPia mater
CArachnoid granulations
DPineal gland
2.Human anatomyHard
After axillary lymph node surgery, a patient's scapula sticks out from the chest wall when pushing against a wall. Which nerve has been damaged?
ASuprascapular nerve
BAxillary nerve
CLong thoracic nerve
DThoracodorsal nerve
3.Human physiologyMedium
Why can cardiac muscle not be tetanised like skeletal muscle?
AIt is controlled only by hormones
BIts long refractory period lasts almost as long as the contraction
CIt is not striated
DIt has no calcium in the sarcoplasmic reticulum
4.Human physiologyHard
In long-standing aortic stenosis, how does the left ventricle adapt?
AAtrophy of the muscle due to less work
BThinning of the wall with dilation
CNo change because the valve is not part of the ventricle
DConcentric hypertrophy with a thicker wall
5.MicrobiologyMedium
Why are pregnant women advised to avoid unpasteurised soft cheeses?
AThey may contain Listeria, which can harm the fetus
BThey often contain the hepatitis C virus, which crosses the placenta
CThey may contain tapeworm eggs
DThey contain too much lactose for the fetus
6.MicrobiologyHard
What does an R0 (basic reproduction number) of 3 mean?
AThree doses of vaccine are needed
BThe infection lasts three days
CThree percent of infected people die
DOn average one case infects three others in a fully susceptible population
7.PharmacologyMedium
A patient starting sulfasalazine should be told about which harmless effect?
AOrange-yellow colouring of the urine
BWhite spots on the nails
CBlue colouring of the tongue and gums
DGreen colouring of the sweat
8.PharmacologyHard
The volume of distribution of a drug is very large (for example 500 L in an adult). What does this suggest?
AThe drug is extensively taken up into tissues
BThe drug stays almost entirely in the plasma compartment
CThe drug is quickly excreted by the kidneys
DThe drug cannot be given intravenously
9.Arterial blood gas interpretationMedium
Practice values: Na⁺ 149 mmol/L, glucose 16.6 mmol/L and urea 10.6 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.
Na⁺ 149 mmol/Lglucose 16.6 mmol/Lurea 10.6 mmol/L
Osm = mOsm/kg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 31 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 31 mmol/L
PaCO₂ = mmHg
11.ECG rhythm strip interpretationMedium
Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Bazett’s formula QTc = QT/√RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
12.ECG rhythm strip interpretationHard
Identify the rhythm on this lead II strip. Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1ventricular fibrillation
2asystole
3junctional rhythm
4sinus bradycardia
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 200 patients received a new treatment and 800 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk RR of the event with treatment. Round to two decimal places.
treatment: 6/200control: 56/800
RR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 50% for a disease. A test with sensitivity 63% and specificity 85% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 50%Se = 63%Sp = 85%negative
p′ = %
3Solutions
1.Human anatomy
Answer: A · Choroid plexus
Why this is the answer
The choroid plexus in the ventricles secretes most CSF. CSF is reabsorbed into venous blood through the arachnoid granulations.
Answer
A · Choroid plexus
2.Human anatomy
Answer: C · Long thoracic nerve
Why this is the answer
The long thoracic nerve supplies serratus anterior, which holds the scapula against the chest wall. Damage causes a winged scapula.
Answer
C · Long thoracic nerve
3.Human physiology
Answer: B · Its long refractory period lasts almost as long as the contraction
Why this is the answer
The plateau of the cardiac action potential gives a long refractory period, so the muscle relaxes before it can be excited again. This allows the heart to refill.
Answer
B · Its long refractory period lasts almost as long as the contraction
4.Human physiology
Answer: D · Concentric hypertrophy with a thicker wall
Why this is the answer
Pressure overload makes the ventricular wall thicken (concentric hypertrophy) to keep wall stress normal. Volume overload, as in regurgitation, causes dilation instead.
Answer
D · Concentric hypertrophy with a thicker wall
5.Microbiology
Answer: A · They may contain Listeria, which can harm the fetus
Why this is the answer
Listeria monocytogenes can grow in chilled foods and cross the placenta. It can cause miscarriage, stillbirth or severe neonatal infection.
Answer
A · They may contain Listeria, which can harm the fetus
6.Microbiology
Answer: D · On average one case infects three others in a fully susceptible population
Why this is the answer
R0 is the average number of secondary cases from one case when everyone is susceptible. Herd immunity needs roughly 1 − 1/R0 of the population immune: about 67% for R0 = 3.
Answer
D · On average one case infects three others in a fully susceptible population
7.Pharmacology
Answer: A · Orange-yellow colouring of the urine
Why this is the answer
Sulfasalazine can colour urine and sometimes skin and contact lenses orange-yellow. More serious adverse effects include blood disorders and rash, so blood counts are monitored.
Answer
A · Orange-yellow colouring of the urine
8.Pharmacology
Answer: A · The drug is extensively taken up into tissues
Why this is the answer
A Vd far above total body water means the plasma concentration is low because most of the drug sits in tissues (for example fat or muscle), as with digoxin or amiodarone.
Answer
A · The drug is extensively taken up into tissues
9.Arterial blood gas interpretation
Answer: Osm = 325 mOsm/kg
Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.
Calculated osmolality
Osm = 2·149 + 16.6 + 10.6 = 325.2 ≈ 325 mOsm/kg
Answer
Osm = 325 mOsm/kg
10.Arterial blood gas interpretation
Answer: PaCO₂ = 42.7 mmHg
Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.
Expected PaCO₂
0.7·31 + 21 = 42.7 mmHg
Answer
PaCO₂ = 42.7 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: QTc = 0.485 s
QT and QTc: the QT interval runs from the start of the QRS complex to the end of the T wave. It shortens as the heart rate rises, so it is corrected for rate with the R–R interval in seconds: Bazett QTc = QT/√RR and Fridericia QTc = QT/∛RR. Bazett is the formula most textbooks use; it over-corrects at fast rates, where Fridericia is preferred. Limits differ between references: a QTc above about 0.45 s in men or 0.46 s in women is usually called prolonged, and above 0.50 s markedly prolonged.
Measure
QT = 10 × 0.04 = 0.40 s
RR = 17 × 0.04 = 0.68 s
Bazett’s formula
QTc = 0.40/√0.68 = 0.40/0.8246 = 0.485 s
Answer
QTc = 0.485 s
Above the usual limits for both men (about 0.45 s) and women (about 0.46 s): prolonged. (485 ms)
12.ECG rhythm strip interpretation
Answer: asystole
Ventricular rhythms: ventricular tachycardia is a run of three or more wide (0.12 s or more) complexes in a row at more than 100/min, usually 150–250/min, regular, without related P waves. Ventricular fibrillation is a chaotic, irregular waveform with no identifiable P waves, QRS complexes or T waves and no countable rate. Asystole is an almost flat line with no ventricular activity; a flat line is always confirmed by checking the patient, the leads and the gain and a second lead. In resuscitation guidelines VF and pulseless VT are the shockable rhythms and asystole is not.
Five steps
1. Rate: no countable ventricular rate.
2. Rhythm: none; an almost flat line.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: no QRS complexes.
Interpretation
asystole
Answer
asystole
A flat line is always checked: look at the patient, the leads and the gain, and view a second lead, because a loose electrode or fine VF can look like asystole.
13.Evidence-based practice statistics
Answer: RR = 0.43
Likelihood ratios and risk measures: LR+ = sensitivity/(1 − specificity), LR− = (1 − sensitivity)/specificity. In a trial, EER and CER are the event rates with and without treatment: RR = EER/CER, ARR = CER − EER, RRR = ARR/CER and NNT = 1/ARR, rounded up. In a case-control study only the odds ratio OR = ad/(bc) can be used. An incidence rate = new cases/person-time.
Event rates
EER = 6/200 = 0.03
CER = 56/800 = 0.07
Relative risk
RR = 0.03/0.07 = 0.4286 ≈ 0.43
Answer
RR = 0.43
14.Evidence-based practice statistics
Answer: p′ = 30.3 %
Predictive values depend on prevalence: PPV = Se·p/(Se·p + (1 − Sp)(1 − p)) and NPV = Sp(1 − p)/(Sp(1 − p) + (1 − Se)p). With likelihood ratios: pre-test odds = p/(1 − p), post-test odds = pre-test odds × LR, probability = odds/(1 + odds). For a population of N, true positives = N·p·Se and false positives = N·(1 − p)·(1 − Sp). NNT = 1/(CER × RRR), rounded up.